Question

Two bicycles are traveling along perpendicular roads. Bicycle A is traveling due east at 4 mi/hr, and bicycle B is travelling due north at 6 mi/hr. At noon, when bicycle A reaches the intersection, bicycle B is 9 mi away and moving toward the same intersection If t is the number of hours after noon, the bicycles are closest together when t isA. 0B. 27/26C. 9/5D. 3/2E. 14/13

Answer 1

Answer:

Step-by-step explanation:

Given

speed of cyclist A is $v_a=4 mi/hr$

speed of cyclist B is $v_b=6 mi/hr$

At noon cyclist B is 9 mi away

after noon Cyclist B will travel a distance of 6 t and cyclist A travel 4 t miles in t hr

Now distance of cyclist B from intersection is 9-6t

Distance of cyclist A from intersection is 4 t

let distance between them be z

$z^2=(9-6t)^2+(4t)^2$

Differentiate z w.r.t time

$2z\frac{\mathrm{d} z}{\mathrm{d} t}=2\times (9-6t)\times (-6)+2\times (4t)\times 4$

$z\frac{\mathrm{d} z}{\mathrm{d} t}=(-6)(9-6t)+4(4t)$

$\frac{\mathrm{d} z}{\mathrm{d} t}=\frac{16t+36t-54}{z}$

Put $\frac{\mathrm{d} z}{\mathrm{d} t}\ to\ get\ maximum\ value\ of\ z$

therefore $52t-54=0$

$t=\frac{54}{52}$

$t=\frac{27}{26} hr$